## Similar Triangles | Introduction

Similar figures: Geometric figures which have the same shape but different sizes are known as similar figures.

**Illustrations:
**

1. Any two line-segments are similar

2. Any two squares are similar

3. Any two circles are similar

Two congruent figures are always similar but two similar figures need not be congruent.

**Similar polygons: **Two polygons of the same number of sides are said to be similar if

(i) Their corresponding angles are equal (i.e., they are equiangular) and

(ii) Their corresponding sides are in the same ratio (or proportion)

**Similar triangles: **Since triangles are also polygons, the same conditions of similarity are applicable to them.

Two triangles are said to be similar if

(i) Their corresponding angles are equal and

(ii) Their corresponding sides are in the same ratio (or proportion).

## BASIC-PROPORTIONALITY THEOREM (Thales theorem)

**Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
**

**Given:** A triangle *ABC *in which a line parallel to side *BC *intersects other two sides *AB *and *AC *at *D *and *E *respectively

**To prove:**

**Construction: **Join *BE *and *CD *and draw *DM *^ *AC *and *EN *^ *AB*.

**Proof:** area of D*ADE*

(Taking *AD* as base)

So,

[The area of D*ADE *is denoted as ar (*ADE*)]

Similarly,

and

(Taking *AE* as base)

Therefore,

…(i)

and

…(ii)

… (iii)

D*BDE *and *DEC *are on the same base *DE *and between the same parallels *BC *and *DE*.]

Therefore, from (i), (ii) and (iii), we have:

**Corollary:
**

From above equation we have

Adding ‘1’ to both sides we have

Þ

**Theorem: (Converse of BPT theorem) If a line divides any two sides of a triangle in the same ratio, prove that it is parallel to the third side.
**

**Given:
**

In D*ABC*, *DE *is a straight line such that

**To prove:
**

*DE* || *BC*

**Construction:
**

If *DE *is not parallel to *BC*, draw *DF* meeting AC at *F*

**Proof:
**

In D*ABC*, let *DF *|| *BC
*

\ …(i)

[ A line drawn parallel to one side of a D divides the other two sides in the same ratio.]

But …(ii) [given]

From (i) and (ii), we get

Adding 1 to both sides, we get

Þ

Þ Þ

*FC* = *EC*

It is possible only when *E* and *F* coincide

Hence, *DE *|| *BC*.

### Question:

*M* and *N* are points on the sides *PQ *and *PR *respectively of D*PQR*. State whether *MN *II *QR*. Given *PQ *= 15.2 cm, *PR *= 12.8 cm, *PM *= 5.7 cm, *PN *= 4.8 cm.

### Solution:

It has been given that

*PQ *= 15.2 cm, *PR *= 12.8 cm,

*PM *= 5.7 cm and *PN *= 4.8 cm

\ *MQ* = *PQ* – *PM* = (15.2 – 5.7) cm = 9.5 cm and

*NR *=* PR – PN *= (12.8 – 4.8) cm = 8 cm

Now

and

\

Thus, in D*PQR*, *MN* divides the sides *PQ* and *PR* in the same ratio. Therefore, by the converse of the Basic Proportionality Theorem, we have *MN *II *QR*.

### Question:

#### In the following figure, if *AB *II *DC*, find the value of *x*.

### Solution:

Since the diagonals of a trapezium divide each other proportionally

\

Þ

Þ

Þ

Þ

Þ

Þ 3*x*(*x* – 2) + 4 (*x* – 2) = 0

Þ

Þ Either or

Þ or

rejected, as it makes line-segments negative]

Thus,

### Question:

#### In the given figure (i) and (ii) *DE* || *BC*. Find *EC* in (i) and *AD* in (ii)

### Solution:

**(i)
**

In D*ABC*, *DE *|| *BC* then by Basic Proportionality (BPT) Theorem, we have

* *or cm

**(ii)
**

Also in figure (ii), in D*ABC, DE *|| *BC *then by B.P.T, we have

or = 2.4 cm

### Question:

#### In the following figure, if *LM* || *CB* and *LN* || *CD*, prove that

### Solution:

**Given:
**

*LM* || *CB* and *LN* || *CD*

**To prove:
**

**Proof:
**

In D*ABC*, *LM* || *BC*, then

By BPT Theorem, we have

…(i) [Corollary of B.P.T.]

Similarly, in D*ADC*, *LN* || *CD
*

…(ii) [Corollary of B.P.T.]

From (i) and (ii), we have

### Question:

#### In the figure, if *DE* || *AC* and *DF* || *AE*, prove that .

### Solution:

In D*ABC*, *DE* || *AC* then by B.P.T., we have

* *…(i)

In D*ABE*, *DF* || *AE* then by B.P.T., we have

* *…(ii)

From (i) and (ii), we get

### Question:

#### In the given figure, if *DE* || *OQ* and *DF* || *OR*, prove that *EF* || *QR*.

### Solution:

In D*POQ*, *DE* || *OQ*, then by BPT Theorem, we have

…(i)

Also, in D*POR*, *DF* || *OR*, then by BPT Theorem, we have

…(ii)

From equations, (i) and (ii),

Þ [converse of BPT Theorem]

### Question:

#### In figure, *PQ* || *AB* and *PR* || *AC*, prove that *QR* || *BC*.

### Solution:

In D*POQ*, *PQ* || *AB*, then by B.P.T., we have

* *…(i)

Also in D*POR*, *AC* || *PR*, we have

* *…(ii)

From (i) and (ii), we get

Þ *QR* || *BC* [by converse of B.P.T.]

### Question:

#### Using Basic proportionality theorem, prove that the line drawn through the mid point of one side of a triangle parallel to another side bisects the third side.

### Solution:

**Given: **A D*ABC* in which *D* is the mid point of *AB* and *DE* || *BC* meeting *AC* at *E*.

**To prove:**

*AE* = *CE*.

**Proof: ** In D*ABC*, *DE* || *BC*

\ [BPT Theorem]

But *AD* = *BD* [*D* is the mid point of *AB*]

\ Þ *AE* = *CE*

Hence, *E* is the mid point of *AC*.

### Question:

*ABCD* is a trapezium such that *AB* || *DC*. The diagonals *AC* and *BD* intersect at *O*. Prove that or .

### Solution:

**Given:**

*ABCD* is a trapezium such that *AB* || *DC
*

**To prove:
**

or

**Construction:
**

Through *O* draw *OE* || *CD*.

**Proof:
**

*AB *|| *DC *[given]

and *EO* || *DC* [const.]

Þ *EO* || *AB*

( Two lines parallel to same line are parallel to each other)

Now, in D*ABD*, *EO* || *AB*, then by BPT Theorem we have

…(i)

Also in D*ADC*, *EO* || *DC*, by BPT Theorem, we have

…(ii)

From (i) and (ii), we get

or

### Question:

#### The diagonals of a quadrilateral *ABCD* intersect each other at the point *O* such that . Show that *ABCD* is a trapezium.

### Solution:

**Given:
**

A quadrilateral *ABCD*, whose diagonals *AC* and *BD* intersect at *O* such that

* *or* *

**To prove:
**

*ABCD* is a trapezium.

**Construction:
**

Though *O*, draw a line *OE* parallel to *AB* intersecting *BC* at *E*.

**Proof:
**

In D*ABC*, *OE* || *AB* [By construction]

\ [by BPT Theorem] …(i)

[given] …(ii)

From (i) and (ii), we have

Now, in D*DBC*, we have

Þ *OE* || *DC* [Converse of BPT Theorem]

Now, in quadrilateral *ABCD*, *OE* || *AB* and *OE* || *DC*

Þ

*AB* || *DC*

( Two lines parallel to same line are parallel to each other)

\ Quadrilateral *ABCD* is a trapezium.